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AleksandrR [38]
3 years ago
10

Given the zeros 7, -1, 9 what are the factors

Mathematics
2 answers:
vlada-n [284]3 years ago
5 0
The factors are (x-7), (x+1), and (x-9)
MrRissso [65]3 years ago
4 0
Poop poop poop poop poop
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Oy is collecting rocks. She has 3 jars and 4 buckets. There are 2 rocks in each jar and 3 rocks in each bucket. How many rocks d
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j = # of jars

b = # of buckets

y = total # of rocks

2j + 3b = y (2 rocks per jar, 3 rocks per bucket)

Because Joy has 3 jars you can plug 3 in for "j", and because she has 4 buckets you can plug in 4 for "b"

2(3) + 3(4) = y

6 + 12 = y

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8 years older then your younger sister

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5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g
anygoal [31]

Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{1117}{10} = 111.7

Sum of squares of differences = 642.1

S.D = \sqrt{\frac{642.1}{49}} = 8.44

We are given the following in the question:  

Population mean, μ = 106

Sample mean, \bar{x} = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

H_{0}: \mu = 106\\H_A: \mu > 106

We use one-tailed(right) t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135

Now,

t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833

Since,                  

t_{stat} > t_{critical}

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

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